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Điều kiện $\sin 2x \ne 0.$
PT $\Leftrightarrow 2 \cos2x\cos \frac{\pi}{4}-2 \sin2x\sin \frac{\pi}{4}=\frac{\cos x}{\sin x}-\frac{\sin x}{\cos x}-2$
$\Leftrightarrow \sqrt 2\left ( \cos2x-\sin 2x \right )=\frac{\cos^2x-\sin^2 x}{\sin x\cos x}-2$
$\Leftrightarrow \sqrt 2\left ( \cos2x-\sin 2x \right )=\frac{2\cos2x}{\sin 2x}-2$
$\Leftrightarrow \sqrt 2\left ( \cos2x-\sin 2x \right )=\frac{2(\cos2x-\sin 2x)}{\sin 2x}$
$\Leftrightarrow \left[ {\begin{matrix} \cos2x=\sin 2x\\ \sin 2x=\frac{1}{\sqrt 2} \end{matrix}} \right.$
$\Leftrightarrow \left[ {\begin{matrix} \cot2x=1\\ \sin 2x=\frac{1}{\sqrt 2} \end{matrix}} \right.$
$\Leftrightarrow \left[ {\begin{matrix} x=\frac{\pi}{8}+k\frac{\pi}{2}\\ x=\frac{\pi}{8}+k\pi\\x=\frac{3\pi}{8}+k\pi \\\end{matrix}} \right. (k \in \mathbb{Z}).$
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